
We apply REF  that \(\mathcal {O}\left(\left\Vert \operatorname{grad}f(\mathbf {P}_t)\alpha _t(\mathbf {P}_t)^\dag \right\Vert \right)\)  has the same scale as \(\mathcal {O}\left(\left\Vert \operatorname{grad}f(\mathbf {P}_t)\right\Vert \right)\) . By difference, it follows
\(\mathbf {P}_{t+1}-\mathbf {P}_t = 4\,\gamma _t \operatorname{sym}\Big (\operatorname{grad}f(\mathbf {P}_t) \alpha _t(\mathbf {P}_t)^\dag \Big )+\mathcal {O}\left(\gamma _t^2\left\Vert \operatorname{grad}f(\mathbf {P}_t)\right\Vert ^2\right).\) 
It implies that there exists a constant \(C_1({X}, r)>0\)  such that
\(\left\Vert  \mathbf {P}_{t+1}-\mathbf {P}_t - 4\,\gamma _t \operatorname{sym}\Big (\operatorname{grad}f(\mathbf {P}_t) \alpha _t(\mathbf {P}_t)^\dag \Big )\right\Vert  \le C_1\gamma _t^2 \left\Vert \operatorname{grad}f(\mathbf {P}_t) \right\Vert ^2.\) 
By REF  in proof difference,
\(\operatorname{grad}f(\mathbf {P}_t) \alpha _t(\mathbf {P}_t)^\dag = (\mathbf {I}_n - \mathbf {P}_t)w(\mathbf {P}_t) \mathbf {P}_t \alpha _t(\mathbf {P}_t)^\dag = (\mathbf {I}_n - \mathbf {P}_t)w(\mathbf {P}_t) \alpha _t(\mathbf {P}_t)^\dag ,\) 
and so
\(\begin{array}{l}\left\langle \operatorname{grad}f(\mathbf {P}_t) \alpha _t(\mathbf {P}_t)^\dag , \alpha _t(\mathbf {P}_t)^\dag \operatorname{grad}f(\mathbf {P}_t) \right\rangle \\=\text{trace}\left( (\mathbf {I}_n - \mathbf {P}_t)w(\mathbf {P}_t) \alpha _t(\mathbf {P}_t)^\dag (\mathbf {I}_n - \mathbf {P}_t)w(\mathbf {P}_t) \alpha _t(\mathbf {P}_t)^\dag \right)=0.\end{array}\) 
It implies
\(\left\Vert \operatorname{sym}\Big (\operatorname{grad}f(\mathbf {P}_t) \alpha _t(\mathbf {P}_t)^\dag \Big )\right\Vert  = \displaystyle \frac{\sqrt{2}}{2}\left\Vert \operatorname{grad}f(\mathbf {P}_t) \alpha _t(\mathbf {P}_t)^\dag \right\Vert .\) 
By the triangle inequality,
\(\begin{array}{ll}\left\Vert  \mathbf {P}_{t+1}-\mathbf {P}_t \right\Vert & \displaystyle \ge 4\times \frac{\sqrt{2}}{2}\gamma _t \left\Vert \operatorname{grad}f(\mathbf {P}_t) \alpha _t(\mathbf {P}_t)^\dag \right\Vert  - C_1\gamma _t^2 \left\Vert \operatorname{grad}f(\mathbf {P}_t) \right\Vert ^2\\& \ge \displaystyle 2\sqrt{2} \times \frac{\sqrt{2}}{3}\gamma _t\left\Vert \operatorname{grad}f(\mathbf {P}_t) \right\Vert  - C_1\gamma _t^2 \left\Vert \operatorname{grad}f(\mathbf {P}_t) \right\Vert ^2\\& = \displaystyle \frac{4}{3}\gamma _t\left\Vert \operatorname{grad}f(\mathbf {P}_t) \right\Vert  - C_1\gamma _t^2 \left\Vert \operatorname{grad}f(\mathbf {P}_t) \right\Vert ^2.\end{array}\) 
Here, the second inequality is due to the first half of REF .
We further constrain the step size as follows
\(0< \gamma _t \le \Gamma _2 :=\min \left(\Gamma _1, \frac{\frac{2}{3}}{C_1\sup _{\mathbf {P}\in \operatorname{Gr}(n,r)}\Vert  \operatorname{grad}f(\mathbf {P})\Vert }\right)\!,\) 
so that
\(C_1\gamma _t^2 \left\Vert \operatorname{grad}f(\mathbf {P}_t) \right\Vert ^2 \le \frac{2}{3} \gamma _t\left\Vert \operatorname{grad}f(\mathbf {P}_t)\right\Vert .\) 
Note that \(\sup _{\mathbf {P}\in \operatorname{Gr}(n,r)} \Vert  \operatorname{grad}f(\mathbf {P}) \Vert \)  is finite because \(\operatorname{grad}f\)  is continuous on \(\operatorname{Gr}(n,r)\)  and the Grassmann is compact. (We also exclude the trivial case when \(\operatorname{grad}f(\mathbf {P})\equiv 0\) .) Continuing from REF  and substituting REF , we have
\(\left\Vert  \mathbf {P}_{t+1}-\mathbf {P}_t \right\Vert  \ge (\frac{4}{3}-\frac{2}{3}) \gamma _t \left\Vert \operatorname{grad}f(\mathbf {P}_t) \right\Vert  = \frac{2}{3}\gamma _t \left\Vert \operatorname{grad}f(\mathbf {P}_t) \right\Vert .\) 
Letting \(\kappa = \inf _{t\rightarrow \infty }\frac{2}{3}\gamma _t > 0\)  ensures that C1 REF  holds.
In preparation for C2, we derive an upper bound for \(\left\Vert  \mathbf {P}_{t+1}-\mathbf {P}_t \right\Vert \) . The idea is similar to REF :
\(\begin{array}{ll}\left\Vert  \mathbf {P}_{t+1}-\mathbf {P}_t \right\Vert  &\le 2\sqrt{2}\gamma _t \left\Vert \operatorname{grad}f(\mathbf {P}_t) \alpha _t(\mathbf {P}_t)^\dag \right\Vert  + C_1\gamma _t^2 \left\Vert \operatorname{grad}f(\mathbf {P}_t) \right\Vert ^2\\&\displaystyle \le 2\sqrt{2} \times \frac{\sqrt{2r}}{2}\gamma _t \left\Vert \operatorname{grad}f(\mathbf {P}_t)\right\Vert +\frac{2}{3}\gamma _t \left\Vert \operatorname{grad}f(\mathbf {P}_t) \right\Vert  \\&\displaystyle \le \frac{8\sqrt{r}}{3}\gamma _t\left\Vert \operatorname{grad}f(\mathbf {P}_t)\right\Vert .\end{array}\) 
The second inequality is obtained from the latter half of REF  and REF .

Keep the assumption REF  on \(\gamma _t\) .
Consider the first-order Taylor's expansion for \(f\)  on \(\operatorname{Gr}(n,r)\)  [1]}.
Inserting REF , the difference of the current and next \(f\)  is:
\(\begin{array}{ll}f(\mathbf {P}_{t+1}) -f(\mathbf {P}_t) & = \langle \operatorname{grad}f(\mathbf {P}_t), \mathbf {P}_{t+1} - \mathbf {P}_t\rangle + \mathcal {O}\left(\Vert  \mathbf {P}_{t+1} - \mathbf {P}_t\Vert ^2\right)\\& = \left\langle \operatorname{grad}f(\mathbf {P}_t),4\gamma _t\, \operatorname{sym}\Big (\operatorname{grad}f(\mathbf {P}_t) \alpha _t(\mathbf {P}_t)^\dag \Big )\right\rangle \\&~~+\mathcal {O}\left(\gamma _t^2\max \left(\left\Vert \operatorname{grad}f(\mathbf {P}_t) \right\Vert ^3, \left\Vert \operatorname{grad}f(\mathbf {P}_t) \right\Vert ^2\right)\right).\end{array}\) 
The first term in the result of REF , since \(\operatorname{grad}f(\mathbf {P}_t)\)  is symmetric, is
\(\begin{array}{l}\left\langle \operatorname{grad}f(\mathbf {P}_t),4\gamma _t \operatorname{sym}\Big (\operatorname{grad}f(\mathbf {P}_t) \alpha _t(\mathbf {P}_t)^\dag \Big )\right\rangle \\= 4\gamma _t \left\langle \operatorname{grad}f(\mathbf {P}_t),\operatorname{grad}f(\mathbf {P}_t) \alpha _t(\mathbf {P}_t)^\dag \right\rangle \\= 4\gamma _t \left\langle \operatorname{grad}f(\mathbf {P}_t)\mathbf {P}_t ,\operatorname{grad}f(\mathbf {P}_t) \alpha _t(\mathbf {P}_t)^\dag \mathbf {P}_t\right\rangle \\= 4\gamma _t \left\langle \operatorname{grad}f(\mathbf {P}_t) \alpha _t(\mathbf {P}_t)^\dag \alpha _t(\mathbf {P}_t),\operatorname{grad}f(\mathbf {P}_t) \alpha _t(\mathbf {P}_t)^\dag \mathbf {P}_t\right\rangle .\end{array}\) 
The second equation is due to \(\alpha _t(\mathbf {P}_t)^\dag \mathbf {P}_t = \alpha _t(\mathbf {P}_t)^\dag .\) 
The definition REF  gives \(\alpha _t(\mathbf {P}_t) = \mathbf {P}_t (\mathbf {I}_n + 2\gamma _t w(\mathbf {P}_t)) \mathbf {P}_t\) . Continuing from above,
\(\begin{array}{l}\left\langle \operatorname{grad}f(\mathbf {P}_t),4\gamma _t \operatorname{sym}\Big (\operatorname{grad}f(\mathbf {P}_t) \alpha _t(\mathbf {P}_t)^\dag \Big )\right\rangle \\= 4\gamma _t \left\langle \operatorname{grad}f(\mathbf {P}_t) \alpha _t(\mathbf {P}_t)^\dag \mathbf {P}_t (\mathbf {I}_n + 2w(\mathbf {P}_t)),\operatorname{grad}f(\mathbf {P}_t) \alpha _t(\mathbf {P}_t)^\dag \mathbf {P}_t\right\rangle \\\ge 4\gamma _t \left\Vert  \operatorname{grad}f(\mathbf {P}_t) \alpha _t(\mathbf {P}_t)^\dag \mathbf {P}_t \right\Vert ^2 = 4\gamma _t \left\Vert  \operatorname{grad}f(\mathbf {P}_t) \alpha _t(\mathbf {P}_t)^\dag \right\Vert ^2\\\ge \displaystyle \frac{8}{9} \gamma _t\Vert  \operatorname{grad}f(\mathbf {P}_t)\Vert ^2.\end{array}\) 
Here we used that \(\mathbf {I}_n + 2\gamma _t w(\mathbf {P}_t) \succeq \mathbf {I}_n\)  for \(\gamma _t \ge 0\) , since \(w(\mathbf {P}_t) \in \mathbb {S}^2(\mathbb {R}^n)\)  is PSD.
The second inequality is derived from REF .
Considering the remainder term in REF , there is a constant \(C_2({X},r)>0\)  such that
\(f(\mathbf {P}_{t+1}) -f(\mathbf {P}_t) \ge \frac{8}{9}\gamma _t\left\Vert  \operatorname{grad}f(\mathbf {P}_t) \right\Vert ^2-C_2\gamma _t^2 \max \left(\left\Vert  \operatorname{grad}f(\mathbf {P}_t)\right\Vert ^3,\left\Vert \operatorname{grad}f(\mathbf {P}_t) \right\Vert ^2 \right).\) 
If the step size \(\gamma _t\)  is small enough to satisfy
\( \begin{array}{ll}0< \gamma _t \!\!\! & \le \Gamma ^* := \min \left( \Gamma _2, \frac{4}{9C_2}\right),\end{array}\) 
where \(\Gamma _2>0\)  is defined in REF , then
\(C_2\gamma _t^2 \max \left(\left\Vert \operatorname{grad}f(\mathbf {P}_t) \right\Vert ^3,\left\Vert \operatorname{grad}f(\mathbf {P}_t)\right\Vert ^2 \right) \le \frac{4}{9}\gamma _t \left\Vert ( \operatorname{grad}f(\mathbf {P}_t)\right\Vert ^2.\) 
Inserting into REF ,
\(\begin{array}{ll}f(\mathbf {P}_{t+1}) -f(\mathbf {P}_t) &\displaystyle \ge \left(\frac{8}{9}-\frac{4}{9}\right)\gamma _t \left\Vert  \operatorname{grad}f(\mathbf {P}_t)\right\Vert ^2\\& = \displaystyle \frac{1}{6\sqrt{r}} \left(\frac{8\sqrt{r}}{3}\gamma _t \left\Vert \operatorname{grad}f(\mathbf {P}_t) \right\Vert \right) \left\Vert \operatorname{grad}f(\mathbf {P}_t) \right\Vert \\[1em]& \ge \displaystyle \frac{1}{6\sqrt{r}}\, \Vert  \mathbf {P}_{t+1} - \mathbf {P}_t\Vert  \Vert \operatorname{grad}f(\mathbf {P}_t)\Vert .\end{array}\) 
The final inequality is by REF .
Thus C2 holds if we set \(\sigma = \frac{1}{6\sqrt{r}}\)  in REF .

“\(\Rightarrow \) ”: From \(\operatorname{grad}f(\mathbf {P}) = \bf 0\)  and REF , we know
\(\begin{array}{ll}\operatorname{grad}f(\mathbf {P})\mathbf {Q}= {\bf 0} =\operatorname{grad}F(\mathbf {Q}) = (\mathbf {I}_n-\mathbf {Q}\mathbf {Q}^{\top })\nabla F(\mathbf {Q}),\end{array}\) 
where \(\mathbf {Q}\in \operatorname{St}(n,r)\)  is a basis of \(\mathbf {P}\in \operatorname{Gr}(n,r)\) . It implies that \(\text{colspan}(\nabla F(\mathbf {Q}))\)  is in \(\text{colspan}(\mathbf {Q}).\)  Note that \(\mathbf {Q}+ \gamma \nabla F(\mathbf {Q}) = \left(\mathbf {I}_n+2\gamma w(\mathbf {P})\right)\mathbf {Q}\)  has full column rank, like \(\mathbf {Q}\) , because \(\mathbf {I}_n+2\gamma w(\mathbf {P}) \in \mathbb {S}^2(\mathbb {R}^n)\)  is positive definite. Hence \(\text{colspan}\left(\mathbf {Q}+ \gamma \nabla F(\mathbf {Q})\right) = \text{colspan}(\mathbf {Q})\) . Then the next iterate \(\Phi (\mathbf {Q})\)  obtained by QR decomposition still outputs a basis for \(\text{colspan}(\mathbf {Q}).\)  So the orthogonal projectors remain the same, i.e. \(\phi (\mathbf {P}) = \mathbf {P}\) .
“\(\Leftarrow \) ”: This directly follows from C1 if \(\gamma _t\)  is sufficiently small to satisfy REF .

Note the Grassmannian is a real-analytic submanifold of \(\operatorname{Gr}(n,r)\) , because \(\operatorname{Gr}(n,r)\)  satisfies property \((3)\)  in [2]}. Also, the cost function \(f\)  REF  is a polynomial in \(\mathbf {P}\in \operatorname{Gr}(n,r)\) , so real-analytic as well. By [3]}, all points \(\mathbf {P}\in \operatorname{Gr}(n,r)\)  satisfy a Łojasiewicz inequality for \(f\) .

